Explanation Given: The differential equation is d y d x = y ln ( y + 2 ) . Calculation: Consider the given differential equation, d y d x = y ln ( y + 2 ) (1) Write the equation of equilibrium and solve for variable y , Substitute 0 for d y d x in equation (1) and solve to find the critical points, y ln ( y + 2 ) = 0 Further solve for factor y = 0 , y = 0 And for factor ln ( y + 2 ) = 0 , ln ( y + 2 ) = 0 y + 2 = 1 y = 1 − 2 y = − 1 Therefore, the critical points are − 1 and 0 . Take the critical points as − 1 and 0 on a vertical line and then defined it into three intervals, as follows, − ∞ < y < − 1 , − 1 < y < 0 and 0 < y < ∞ . Write the differential equation in terms of f ( y ) as follows, f ( y ) = y ln ( y + 2 ) (2) To find the algebraic sign take the points on the constructed intervals and substitute it in equation (2) as below, For interval − ∞ < y < − 1 substitute − 1.5 for y in equation (2), f ( − 1.5 ) = ( − 1.5 ) ln ( − 1.5 + 2 ) = ( − 1.5 ) ln ( 0.5 ) = 1.0397 Therefore for interval − ∞ < y < − 1 the algebraic sign is positive. For interval − 1 < y < 0 substitute − 0.5 for y in equation (2), f ( − 0

Differential Equations with Boundary-Value Problems (MindTap Course List)

9th Edition

ISBN: 9781305965799

Author: Dennis G. Zill

Publisher: Cengage Learning

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Differential Equations

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Chapter 2.1, Problem 27E

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To sketch: the typical solution curve in the region in the xy-plane determined by the graph of equilibrium solutions.

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Chapter 2.1, Problem 26E

Chapter 2.1, Problem 28E

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Chapter 2 Solutions

Differential Equations with Boundary-Value Problems (MindTap Course List)

Ch. 2.1 - In Problems 14 reproduce the given...Ch. 2.1 - In Problems 14 reproduce the given...Ch. 2.1 - dydx=1xy (a) y(0) = 0 (b) y(1) = 0 (c) y(2) = 2...Ch. 2.1 - In Problems 14 reproduce the given...Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E Ch. 2.1 - Prob. 7E Ch. 2.1 - Prob. 8E Ch. 2.1 - Prob. 9E Ch. 2.1 - Prob. 10E

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the typical solution curve in the region in the x y -plane determined by the graph of equilibrium solutions.

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